Question

In lab you need to prepare at least \(100 \mathrm{mL}\) of each of the following solutions. Explain how you would proceed using the given information. a. \(2.0 \mathrm{m} \mathrm{KCl}\) in water (density of \(\mathrm{H}_{2} \mathrm{O}=1.00 \mathrm{g} / \mathrm{cm}^{3}\) ) b. \(15 \%\) NaOH by mass in water \(\left(d=1.00 \mathrm{g} / \mathrm{cm}^{3}\right)\) c. \(25 \%\) NaOH by mass in \(\mathrm{CH}_{3} \mathrm{OH}\left(d=0.79 \mathrm{g} / \mathrm{cm}^{3}\right)\) d. 0.10 mole fraction of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) in water \(\left(d=1.00 \mathrm{g} / \mathrm{cm}^{3}\right)\)

Short answer

To prepare the following solutions: a. 2.0 m KCl in water: Weigh 14.91 g of KCl, dissolve in a small amount of water, transfer to a 100 mL volumetric flask, fill to the mark with water, and mix. b. 15% NaOH by mass in water: Weigh 15 g of NaOH, dissolve in a small amount of water, transfer to a 100 mL volumetric flask, fill to the mark with water, and mix. c. 25% NaOH by mass in CH3OH: Weigh 19.75 g of NaOH, dissolve in a small amount of CH3OH, transfer to a 100 mL volumetric flask, fill to the mark with CH3OH, and mix. d. 0.10 mole fraction of C6H12O6 in water: Weigh 111.2 g of C6H12O6, dissolve in a small amount of water, transfer to a 100 mL volumetric flask, fill to the mark with water, and mix.

Step-by-step solution

Calculate the mass of solute needed

For a 2.0 m KCl solution, there should be 2 moles of KCl in 1 kg of water (since molality is defined as moles of solute per kilogram of solvent). To prepare 100 mL of this solution, we first need to find out how much KCl is needed. Since the density of water is given as \(1.00 g/cm^3\), the mass of 100 mL of water is \(100g\). So, we have \(1/10\) kg of water, which means we need \(2/10\) moles of KCl. The molar mass of KCl is \(39.10 + 35.45 = 74.55 g/mol\). Therefore, we need \(\frac{2}{10} \times 74.55 = 14.91 g\) of KCl.

Prepare the solution using the calculated mass of solute

To prepare the solution, first weigh out 14.91 g of KCl. Next, dissolve the KCl in a small amount of water and transfer the solution into a 100 mL volumetric flask. Finally, add water up to the 100 mL mark (make sure the volume of the solution reaches the 100 mL line), and mix the solution thoroughly. b. 15% NaOH by mass in water

Calculate the mass of solute needed

In this case, we have 15% NaOH by mass in the solution. The density of the solution is \(1.00 g/cm^3\). This means we have a total mass of 100 g for 100 mL in which 15% is NaOH. Therefore, we need \(0.15 \times 100 = 15 g\) of NaOH.

Prepare the solution using the calculated mass of solute

To prepare the solution, first weigh out 15 g of NaOH. Next, dissolve the NaOH in a small amount of water and transfer the solution into a 100 mL volumetric flask. Finally, add water up to the 100 mL mark and mix the solution thoroughly. c. 25% NaOH by mass in CH3OH

Calculate the mass of solute needed

In this case, we have a solution of NaOH in CH3OH with a density of \(0.79 g/cm^3\). The mass of 100 mL of CH3OH is \(0.79 \times 100 = 79 g\). Since it's a 25% NaOH solution, we need \(0.25 \times 79 = 19.75 g\) of NaOH.

Prepare the solution using the calculated mass of solute

To prepare the solution, first weigh out 19.75 g of NaOH. Next, dissolve the NaOH in a small amount of CH3OH and transfer the solution into a 100 mL volumetric flask. Finally, add CH3OH up to the 100 mL mark and mix the solution thoroughly. d. 0.10 mole fraction of C6H12O6 in water

Calculate the mass of solute needed

In this case, we have a mole fraction of \(0.10\) for C6H12O6 in water. This means that the ratio of moles of C6H12O6 to the total moles of the solution (including solvent) is \(0.10\). The molar mass of C6H12O6 is \(180.16 g/mol\), and since we know the density of water is \(1.00 g/cm^3\), the mass of 100 mL of water is 100 g. Molar mass of water is \(18 g/mol\). So we have \(100/18 = 5.56\) moles of water. Since the mole fraction of C6H12O6 is \(0.1\), we have \(0.1/0.9\) times the moles of water, which is \(\frac{0.1}{0.9} \times 5.56 = 0.617\) moles of C6H12O6. Therefore, we need \(0.617 \times 180.16 = 111.2 g\) of C6H12O6.

Prepare the solution using the calculated mass of solute

To prepare the solution, first weigh out 111.2 g of C6H12O6. Next, dissolve the C6H12O6 in a small amount of water and transfer the solution into a 100 mL volumetric flask. Finally, add water up to the 100 mL mark and mix the solution thoroughly.

Key concepts

Molality and Molarity

Understanding the concepts of molality and molarity is essential in solution preparation. Molality (\b m \b) refers to the number of moles of solute per kilogram of solvent, and is a temperature-independent measure since mass doesn't change with temperature.

For example, a 2.0 m KCl solution requires 2 moles of KCl dissolved in 1 kg of water. To determine the amount needed for 100 mL of this solution, one would compute the molality considering the volume and density of water. Doing the math, \b \(\frac{2}{10}\) moles of KCl corresponds to about 14.91 g of KCl.

Molarity (\b M \b), on the other hand, is the number of moles of solute per liter of solution. Molarity changes with temperature as the solution's volume may expand or contract. For educational purposes, it is important to distinguish between the two concepts and their applications in different scenarios.

Percent Composition by Mass

Percent composition by mass indicates the mass percentage of each component in a mixture. When the exercise asks for a 15% NaOH solution by mass, it implies that 15 g of NaOH should be present in every 100 g of solution.

Calculating the amount of NaOH needed is straightforward: \b \(0.15 \times 100 = 15 g\) \b. Observing the density of water can guide the preparation process, ensuring that the percentage remains accurate. Similar calculations follow for preparing a solution with any other percentage composition by mass, such as a 25% NaOH solution in methanol, where methanol's density impacts the mass used for calculations.

Mole Fraction

The mole fraction is a way of expressing the ratio of the number of moles of one component to the total number of moles in a mixture. A 0.10 mole fraction of \b C₆H₁₂O₆ \b in water indicates that for every 10 moles of solution, 1 mole is glucose.

The calculation of the required amount of solute is based on the mole ratio and total moles in the solution. For a 0.10 mole fraction in 100 mL of water, one calculates the moles of water and then determines the moles of solute as \b \(\frac{0.1}{0.9} \times 5.56 = 0.617 moles\) \b of glucose, which equals to 111.2 g of glucose to be dissolved. This allows precise preparation of solutions with specific mole fractions.

Volumetric Solution Preparation

Volumetric solution preparation involves measuring and mixing precise amounts of solute and solvent to achieve a desired concentration. It typically involves using a volumetric flask, which allows for accurate volume measurements up to a specific mark.

For instance, after calculating and weighing the solute such as KCl, NaOH, or glucose, it's dissolved in a minimal amount of the solvent. The solution is then transferred to the volumetric flask, and more solvent is added until the bottom of the meniscus touches the calibration line of the flask.

It's crucial to ensure complete dissolution of the solute and homogeneous mixing. Volumetric preparation is fundamental in the lab for creating standards and solutions for reactions or analytical techniques.